Simplify; express your answer in exponential form. Assume $k\neq 0, y\neq 0$. $\dfrac{{(k^{3})^{2}}}{{(k^{-5}y^{-5})^{-5}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{3}}$ to the exponent ${2}$ . Now ${3 \times 2 = 6}$ , so ${(k^{3})^{2} = k^{6}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-5}y^{-5})^{-5} = (k^{-5})^{-5}(y^{-5})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{3})^{2}}}{{(k^{-5}y^{-5})^{-5}}} = \dfrac{{k^{6}}}{{k^{25}y^{25}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{6}}}{{k^{25}y^{25}}} = \dfrac{{k^{6}}}{{k^{25}}} \cdot \dfrac{{1}}{{y^{25}}} = k^{{6} - {25}} \cdot y^{- {25}} = k^{-19}y^{-25}$.